In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
Practice Set 1.3 | Q 9 | Page 22 In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
Solution
Given: ∠BAC = ∠ADC To prove: CA2 = CB × CD Proof: In ∆ABC and ∆DAC ∠BAC = ∠ADC (Given) ∠C = ∠C (Common) By AA test of similarity ∆ABC ∼ ∆DAC
\[\therefore \frac{BC}{AC} = \frac{AC}{DC} \left( \text{ Corresponding sides are proportional } \right)\] \[ \Rightarrow {AC}^2 = BC \times DC\]
Hence proved.
Answer:-
Given: ∠BAC = ∠ADC To prove: CA^2 = CB × CD Proof: In ∆ABC and ∆DAC, we have: ∠BAC = ∠ADC … (Given) ∠C = ∠C … (Common angle) By AA test of similarity, we have: ∆ABC ~ ∆DAC So, we can write the following proportion: BC/AC = AC/DC … (Corresponding sides are proportional) This can be simplified to: AC^2 = BC × DC Hence, we have proved that CA^2 = CB × CD.